Source Latex
de la correction du devoir
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pdfauthor={Yoann Morel},
pdfsubject={Corrigé du devoir de mathématiques: Calcul algébrique, fractions, développement et factorisation},
pdftitle={Corrigé du devoir de mathématiques},
pdfkeywords={calcul algébrique, fraction, développement, factorisation, identités remarquables, mathématiques, seconde, 2nde}
}
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\author{Y. Morel}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{document}
\textbf{\fbox{\Large B}}\hfill{\Large\textbf\TITLE}\hfill \,
\bigskip
\bgex
\[\bgar{ll}
A&=\dfrac{3}{x+2}-\dfrac{1}{x+1}
=\dfrac{3(x+1)}{(x+2)(x+1)}-\dfrac{1(x+2)}{(x+2)(x+1)}\\[1em]
&=\dfrac{3(x+1)-(x+2)}{(x+2)(x+1)}
=\dfrac{2x+1}{(x+2)(x+1)}
\enar\]
\[B=\dfrac{3x}{x+2}-3
=\dfrac{3x}{x+2}-\dfrac{3(x+2)}{x+2}
=\dfrac{3x-3(x+2)}{x+2}=\dfrac{-6}{x+2}\]
\[\bgar{ll}
C&=3+\dfrac{\dfrac13\sqrt3}{1+\dfrac{\sqrt3}{3}}
=3+\dfrac{\dfrac{\sqrt3}{3}}{\dfrac{3+\sqrt3}{3}}
=3+\dfrac{\sqrt3}{3}\tm\dfrac{3}{3+\sqrt3}
=3+\dfrac{\sqrt3}{3+\sqrt3}\\[2.4em]
&=3+\dfrac{\sqrt3(3-\sqrt3)}{(3+\sqrt3)(3-\sqrt3)}
=3+\dfrac{3\sqrt3-3}{3^2-\sqrt3^2}
=3+\dfrac{3(\sqrt3-1)}{6}\\[1.5em]
&=3+\dfrac{\sqrt3-1}{2}=\dfrac{5+\sqrt3}{2}
\enar\]
\[\bgar{ll}
D&=\lp2-\dfrac{5}{\sqrt3+3}\rp^2
=\lp 3-\dfrac{5(\sqrt2-2)}{(\sqrt2+2)(\sqrt2-2)}\rp^2
=\lp 3-\dfrac{5\sqrt2-10}{-2}\rp^2\\[1.6em]
&=\lp\dfrac{3\tm(-2)-\lp5\sqrt2-10\rp}{-2}\rp^2
=\lp\dfrac{4-5\sqrt2}{-2}\rp^2
=\dfrac{\lp4-5\sqrt2\rp^2}{(-2)^2}\\[1.6em]
&=\dfrac{4^2-2\tm4\tm5\sqrt2+\lp5\sqrt2\rp^2}{4}
=\dfrac{16-40\sqrt2+50}{4}
=\dfrac{66-40\sqrt2}{4}\\[1.6em]
&=\dfrac{2\lp33-20\sqrt2\rp}{2\tm2}
=\dfrac{33-20\sqrt2}{2}
\enar\]
\enex
\bgex
\[A=(3x+1)-(4-x)(3x+1)
=(3x+1)\Bigl(1-(4-x)\Bigr)
=(3x+1)(-3+x)\]
\[B=(x+2)(-x+1)-(x+2)^2
=(x+2)\Bigl((-x+1)-(x+2)\Bigr)
=(x+2)(-2x-1)\]
\[C=(3x+1)^2-9=(3x+1)^2-3^2
=\Bigl((3x+1)-3\Bigr)\Bigl((3x+1)+3\Bigr)
=(3x-2)(3x+4)\]
\enex
\label{LastPage}
\end{document}
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